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3 years ago

A cannonball is launched at ground level at an angle of 30° above the horizontal with

Physics

2 answers:

user100 [1]3 years ago

8 0

Answer:

use the kinematics equations and solve for time, after that use dleta dx=Vi*delta time

Nookie1986 [14]3 years ago

8 0

Use the kemetic equation

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How is heat related to phase changes

bekas [8.4K]

The heat is used to break the bonds between the ice molecules as they turn into a liquid phase. Since the average kinetic energy of the molecules does not change at the moment of melting, the temperature of the molecules does not change.

7 0

3 years ago

A 1 kg particle moves upward from the origin to (23) m. Wit is work done by the force of gravity which is in - y direction s B.

sveta [45]

Answer:

Explanation:

mass, m = 1 kg

Position (2, 3 ) m

height, h = 2 m

acceleration due to gravity, g = 9.8 m/s^2

Here, no force is acting in horizontal direction, the force of gravity is acting in vertical direction, so the work done by the gravitational force is to be calculated.

Force mass x acceleration due to gravity

F = 1 x 9.8 = 9.8 N

Work = force x displacement x CosФ

Where, Ф be the angle between force vector and the displacement vector.

Here the value of Ф is 180° as the force acting vertically downward and the displacement is upward

So, W = 9.8 x 2 x Cos 180°

W = - 19.6 J

Thus, option (A) is correct.

4 0

2 years ago

What is the spring constant for a supermarket scale that stretches 0.01 m when a force of 4 N is applied

bija089 [108]

Answer:

400N/m

Explanation:

K = F/x

= 4/0.01

= 400 N/m

5 0

3 years ago

The two pucks of equal mass did not move linearly (they came to a stop) after the collision due to the conservation of linear mo

weeeeeb [17]

Compared to the pucks given, the pair of pucks will rotate at the same rate.

Answer: Option A

<u>Explanation:</u>

The law of conservation of the angular momentum expresses that when no outer torque follows upon an article, no difference in angular momentum will happen. At the point when an item is turning in a shut framework and no outside torques are applied to it, it will have no change in angular momentum.

The conservation of the angular momentum clarifies the angular quickening of an ice skater as she brings her arms and legs near the vertical rotate of revolution. In the event, that the net torque is zero, at that point angular momentum is steady or saved.

By twice the mass yet keeping the speeds unaltered, also twice the angular momentum's to the two-puck framework. Be that as it may, we likewise double the moment of inertia. Since $L=I \times \omega$ , the turning rate of the two-puck framework must stay unaltered.

4 0

3 years ago

Suppose the student in (Figure 1) is 68kg, and the board being stood on has a 12kg mass. What is the reading on the left scale?

lesantik [10]

The equilibrium conditions allow to find the results for the balance forces are:

F₁ = 225.4 N

F₂ = 558.6 N

When the acceleration is zero we have the equilibrium conditions for both linear and rotational motion.

∑ F = 0

∑ τ = 0

Where F are the forces and τ the torques.

The torque is the product of the force and the perpendicular distance to the point of support,

The free-body diagrams are diagrams of the forces without the details of the bodies, see attached for the free-body diagram of the system.

We write the translational equilibrium condition.

F₁ - W₁ - W₂ + F₂ = 0

We write the equation for the rotational motion, set our point of origin at scale 1, and the counterclockwise turns are positive.

F₂ 2 - W₁ 1 - W₂ 1.5 = 0 $\frac{W_1 \ 1 + W_2 \ 1.5}{2}$